What is the equation of the line tangent to the graph of the function f(x)=x2+2x−5 at the point (2,3) ?
To find the equation of the tangent line to the graph of the function f(x) = x^2 + 2x - 5 at the point (2, 3), we can use the point-slope form of the equation of a line.
First, we need to find the slope of the tangent line. Since the tangent line is a line tangent to the graph of the function at a specific point, the slope of the tangent line at that point is equal to the derivative of the function at that point.
Taking the derivative of f(x) = x^2 + 2x - 5 with respect to x, we get:
f'(x) = 2x + 2
Now, we can substitute x = 2 into the derivative function to find the slope at the point (2, 3):
m = f'(2) = 2(2) + 2 = 6
So, the slope of the tangent line at the point (2, 3) is 6.
Next, we can use the point-slope form of the equation of a line to find the equation of the tangent line:
y - y1 = m(x - x1)
Substituting the values y1 = 3, x1 = 2, and m = 6, we get:
y - 3 = 6(x - 2)
Expanding and simplifying:
y - 3 = 6x - 12
Rearranging the equation, we get the equation of the tangent line in slope-intercept form:
y = 6x - 9
So, the equation of the line tangent to the graph of the function f(x) = x^2 + 2x - 5 at the point (2, 3) is y = 6x - 9.
To find the equation of the line tangent to the graph of the function f(x) = x^2 + 2x - 5 at the point (2,3), we need to find the slope of the tangent line and use the point-slope form of a line.
Step 1: Find the derivative of the function.
The derivative of f(x) = x^2 + 2x - 5 is given by f'(x) = 2x + 2.
Step 2: Find the slope of the tangent line.
Evaluate the derivative at x = 2 to find the slope of the tangent line at that point.
f'(2) = 2(2) + 2 = 4 + 2 = 6.
So, the slope of the tangent line is 6.
Step 3: Use the point-slope form to find the equation of the tangent line.
Using the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the point on the line and m is the slope, substitute the values into the equation.
(x1, y1) = (2, 3) and m = 6.
Therefore, the equation of the line tangent to the graph of f(x) = x^2 + 2x - 5 at the point (2,3) is:
y - 3 = 6(x - 2).
To find the equation of the line tangent to the graph of a function at a given point, you need to determine the slope of the tangent line and the coordinates of the point. The slope of the tangent line is equal to the derivative of the function at that point.
First, let's find the derivative of the function f(x) = x^2 + 2x - 5. Taking the derivative will give us the slope of the tangent line at any point on the graph.
To find the derivative, differentiate each term of the function with respect to x. The derivative of x^2 with respect to x is 2x, the derivative of 2x with respect to x is 2, and the derivative of a constant like -5 is 0. Taking these derivatives, we have:
f'(x) = 2x + 2
Now that we have the derivative, we can determine the slope of the tangent line at the point (2,3) by substituting x=2 into the derivative:
f'(2) = 2(2) + 2 = 4 + 2 = 6
So, the slope of the tangent line at the point (2,3) is 6.
Now that we have the slope of the tangent line and one point (2,3), we can use the point-slope form of the equation of a line to find the equation of the tangent line. The point-slope form is given by:
y - y1 = m(x - x1)
where m is the slope and (x1, y1) is the given point.
Substituting the values we have, we get:
y - 3 = 6(x - 2)
Now, simplify this equation to get it into slope-intercept form, y = mx + b:
y - 3 = 6x - 12
y = 6x - 9
Therefore, the equation of the line tangent to the graph of the function f(x) = x^2 + 2x - 5 at the point (2,3) is y = 6x - 9.